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flts.4
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1999-09-16
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flts(1) Scilab Function flts(1)
NAME
flts - time response (discrete time, sampled system)
CALLING SEQUENCE
[y [,x]]=flts(u,sl [,x0])
[y]=flts(u,sl [,past])
PARAMETERS
u : matrix (input vector)
sl : list (linear system syslin)
x0 : vector (initial state ; default value=0)
past : matrix (of the past ; default value=0)
x,y : matrices (state and output)
DESCRIPTION
State-space form:
sl is a syslin list containing the matrices of the following linear system
sl=syslin('d',A,B,C,D) (see syslin):
x[t+1] = A x[t] + B u[t]
y[t] = C x[t] + D u[t]
or, more generally, if D is a polynomial matrix (p = degree(D(z))) :
D(z)=D_0 + z D_1 + z^2 D_2 +..+ z^p D_p
y[t] = C x[t] + D_0 u[t] + D_1 u[t+1] +..+ D_[p] u[t+p]
u=[u0,u1,... un] (input)
y=[y0,y1,... yn-p] (output)
x=x[n-p+1] (final state, used as x0 at next call to flts)
Transfer form:
y=flts(u,sl[,past]). Here sl is a linear system in transfer matrix
representation i.e
sl=syslin('d',transfer_matrix) (see syslin).
past = [u ,..., u ]
[ -nd -1]
[y ,..., y ]
[ -nd -1]
is the matrix of past values of u and y.
nd is the maximum of degrees of lcm's of each row of the denominator matrix
of sl.
u=[u0 u1 ... un] (input)
y=[y0 y1 ... yn] (output)
p is the difference between maximum degree of numerator and maximum degree
of denominator
EXAMPLE
sl=syslin('d',1,1,1);u=1:10;
y=flts(u,sl); [y1,x1]=flts(u(1:5),sl);y2=flts(u(6:10),sl,x1);
y-[y1,y2]
//With polynomial D:
z=poly(0,'z');sl=syslin('d',1,1,1,1+z+z^2);p=2;
y=flts(u,sl);[y1,x1]=flts(u(1:5),sl);
y2=flts(u(5-p+1:10),sl,x1); // (update)
y-[y1,y2]
//Delay (transfer form): flts(u,1/z)
SEE ALSO
ltitr, dsimul, rtitr